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Math Help - Integrating absolute values (definite integrals) - can't find good info

  1. #1
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    Integrating absolute values (definite integrals) - can't find good info

    Anyone have a good tutorial? Google just sends me to people asking questions like I am. I know it's supposed to be split up. Anyways, here is the problem I am stuck on:

    from -2 to 2 integrate (1 - 3|x|)dx

    Whenever I think about splitting it up on a definite integral, my brain sort of melts. My book is no help either
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  2. #2
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    Yes, when dealing with absolute values, break them up.

    All it is is two lines. One for negative and one for positive.

    1-3|x| is 1+3x and 1-3x

    \int_{-2}^{0}(1+3x)dx+\int_{0}^{2}(1-3x)dx
    Attached Thumbnails Attached Thumbnails Integrating absolute values (definite integrals) - can't find good info-abs1.jpg  
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  3. #3
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    Quote Originally Posted by galactus View Post
    Yes, when dealing with absolute values, break them up.

    All it is is two lines. One for negative and one for positive.

    1-3|x| is 1+3x and 1-3x

    \int_{-2}^{0}(1+3x)dx+\int_{0}^{2}(1-3x)dx
    clear, concise

    thank you
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  4. #4
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    Hello, TYTY!

    With absolute values, a sketch usually helps . . .


    \int^2_{-2}\left(1-3|x|\right)\,dx

    We know the graph of y = |x| . . . a V with vertex at the origin.

    The graph of y = 3|x| . . . "climbs steeper".

    The graph of y \:=\:-3|x| . . . inverts the previous graph.

    The graph of y \:=\:1-3|x| . . . moved up one unit.
    Code:
                  |
                  *
                 /|\
                /:|:\
         -2    /::|::\    2
      - - + - o - + - o - + - -
          |::/    |    \::|
          |:/     |     \:|
          |/      |      \|
          *       |       *
                  |

    We have an inverted V, vertex at (0,1), x-intercepts at \pm\tfrac{1}{3}

    They ask for the "area" from x = \text{-}2 to x=2,
    . . and some of it is "negative area".

    We don't need Calculus to find the areas of triangles.

    [My answer is: -8]

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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, TYTY!

    With absolute values, a sketch usually helps . . .



    We know the graph of y = |x| . . . a V with vertex at the origin.

    The graph of y = 3|x| . . . "climbs steeper".

    The graph of y \:=\:-3|x| . . . inverts the previous graph.

    The graph of y \:=\:1-3|x| . . . moved up one unit.
    Code:
                  |
                  *
                 /|\
                /:|:\
         -2    /::|::\    2
      - - + - o - + - o - + - -
          |::/    |    \::|
          |:/     |     \:|
          |/      |      \|
          *       |       *
                  |

    We have an inverted V, vertex at (0,1), x-intercepts at \pm\tfrac{1}{3}

    They ask for the "area" from x = \text{-}2 to x=2,
    . . and some of it is "negative area".

    We don't need Calculus to find the areas of triangles.

    [My answer is: -8]

    This is a good point and also how the text wanted it to be done. When I thought about integrating it is where I tripped up though.
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